I'm wondering if there is any way to prove that of the numbers in the form 1.....1 (i.e. all digits are 1), 1 is the only cube number.
i was able to narrow it down a bit. i could prove that the only candidates for being a cube are the numbers of the form $\displaystyle (10^{6m+1}-1)/9, \, (10^{6m+3}-1)/9, \, m \in \mathbb{Z}^+$.
i can post the proof of the above if you are familiar with modular arithmetic and fermat's little theorem.
first we use the fact that $\displaystyle x^3 \equiv 0,\, 1, \, -1\,(\,mod\,7\,)$.
proof: using fermat's li'l theorem $\displaystyle x^6 \equiv 0, \,1 $ so either $\displaystyle x \equiv 0 \, (\, mod \, 7 \, )$ or $\displaystyle x^6 \equiv 1 \,(\, mod \, 7 \,)$. the latter implies $\displaystyle (x^3-1)(x^3+1) \equiv 0 \,(\,mod \, 7 \,)\Rightarrow x^3 \equiv 1,\,-1 \,(\,mod \, 7\,)$.
so if $\displaystyle (10^y-1)/9$ has to be a cube then the inly admissible values of $\displaystyle y$ would be the ones satisfying $\displaystyle (10^y-1)/9 \equiv 0,\,1\,-1\,(\,mod\,7\,)$. there are only $\displaystyle 7$ trials needed to know which are those. the possible values of $\displaystyle y$ are $\displaystyle y=6m,\,6m+1,\,6m+3$. that's it.
in my last post i missed the numbers of the type $\displaystyle (10^{6m}-1)/9$.